Shooting Method On Harmonic Equation

2 vues (au cours des 30 derniers jours)
Alexander Kimbley
Alexander Kimbley le 19 Fév 2019
Modifié(e) : Torsten le 28 Fév 2019
I'm really new to Matlab, so this may be ridicously easy what I'm about to ask, but bare with me please.
I'm trying to integrate the Harmonic equation y'' +(a^2)*y=0 with a=2.4, with BC y(0)=y'(pi)=0. I'm doing this to try and "shoot" for the actual value of y at y=pi which we obviously can find analytically but I need to get my head around the code so I can apply this to a more complicated problem.
Thanks.

Réponse acceptée

Torsten
Torsten le 20 Fév 2019
function main
ydot0_start = 1.0;
a = 2.4;
iflag = 0;
sol = fzero(@(x)fun_shooting(x,a,iflag),ydot0_start);
iflag = 1;
y_at_pi = fun_shooting(sol,a,iflag)
end
function res = fun_shooting(x,a,iflag)
fun_ode = @(t,y)[y(2);-a^2*y(1)];
tspan = [0,pi];
y0 = [0;x];
[t,y] = ode45(fun_ode,tspan,y0);
if iflag == 0
res = y(end,2);
else
res = y(end,1);
end
end
  8 commentaires
Alexander Kimbley
Alexander Kimbley le 27 Fév 2019
So how would I go about changing the conditions to y(0)=0, y'(0)=1, where we alter 'a' to get y(pi)=0 to 10^-8 accuaracy say, assuming we dont actually know the true value of a.
Thanks.
Torsten
Torsten le 28 Fév 2019
Modifié(e) : Torsten le 28 Fév 2019
function main
a0 = 4.5;
a = fzero(@fun_shooting,a0)
end
function res = fun_shooting(x)
fun_ode = @(t,y)[y(2);-x^2*y(1)];
tspan = [0,pi];
y0 = [0;1];
[t,y] = ode45(fun_ode,tspan,y0);
res = y(end,1);
end

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